Problem: This is a harder version of the problem. In this version q ≤ 200 000.

A sequence of integers is called nice if its elements are arranged in blocks like in [3, 3, 3, 4, 1, 1]. Formally, if two elements are equal, everything in between must also be equal.

Let's define difficulty of a sequence as a minimum possible number of elements to change to get a nice sequence. However, if you change at least one element of value x to value y, you must also change all other elements of value x into y as well. For example, for [3, 3, 1, 3, 2, 1, 2] it isn't allowed to change first 1 to 3 and second 1 to 2. You need to leave 1's untouched or change them to the same value.

You are given a sequence of integers a_1, a_2, …, a_n and q updates.

Each update is of form "i x" — change a_i to x. Updates are not independent (the change stays for the future).

Print the difficulty of the initial sequence and of the sequence after every update.

Input

The first line contains integers n and q (1 ≤ n ≤ 200 000, 0 ≤ q ≤ 200 000), the length of the sequence and the number of the updates.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000), the initial sequence.

Each of the following q lines contains integers i_t and x_t (1 ≤ i_t ≤ n, 1 ≤ x_t ≤ 200 000), the position and the new value for this position.

Output

Print q+1 integers, the answer for the initial sequence and the answer after every update.

Example

Input


5 6
1 2 1 2 1
2 1
4 1
5 3
2 3
4 2
2 1


Output


2
1
0
0
2
3
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void maxtt(T &t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T &t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
string direc = "URDL";
const long long MOD2 = (long long)1000000007 * (long long)1000000007;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int a[200135], Q;
set<int> pos[200135];
const int MN = (1 << 20) + 5;
struct mb {
  int M, T[MN];
  void init(int sz) {
    for (M = 1; sz + 2 > M; M <<= 1)
      ;
  }
  void init(int sz, int val) {
    for (M = 1; sz + 2 > M; M <<= 1)
      ;
    for (int(i) = 0; (i) < (int)(MN); (i)++) T[i] = val;
  }
  void clear() { memset(T, 0, sizeof T); }
  void upt(int l, int val) {
    l += M;
    T[l] = val;
    while (l > 1) {
      T[l >> 1] = max(T[l], T[l ^ 1]);
      l >>= 1;
    }
  }
  int qyRangeMax(int l, int r) {
    if (r < l) return 0;
    int ans = 0;
    for (l = l + M - 1, r = r + M + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
      if (~l & 1) ans = max(ans, T[l ^ 1]);
      if (r & 1) ans = max(ans, T[r ^ 1]);
    }
    return ans;
  }
} T1;
struct nd {
  int lp, rp, mi, sum;
  nd() { lp = rp = mi = sum = 0; }
  friend nd operator+(const nd &x, const nd &y) {
    nd res = nd();
    if (x.mi < y.mi)
      res = x;
    else if (y.mi < x.mi)
      res = y;
    else {
      res.mi = x.mi;
      res.lp = x.lp;
      res.rp = y.rp;
      res.sum = x.sum + y.sum + T1.qyRangeMax(x.rp, y.lp - 1);
    }
    return res;
  }
  void addlz(int val) { mi += val; }
} T[200135 * 4];
int lz[200135 * 4];
void Up(int rt, int l, int r) { T[rt] = T[(rt << 1)] + T[(rt << 1 | 1)]; }
void Down(int rt, int l, int r) {
  if (lz[rt] != 0) {
    T[(rt << 1)].addlz(lz[rt]);
    lz[(rt << 1)] += lz[rt];
    T[(rt << 1 | 1)].addlz(lz[rt]);
    lz[(rt << 1 | 1)] += lz[rt];
    lz[rt] = 0;
  }
}
void build(int rt, int l, int r) {
  if (l == r) {
    T[rt].lp = r;
    T[rt].rp = r;
    return;
  }
  int mid = (l + r) / 2;
  build((rt << 1), l, mid);
  build((rt << 1 | 1), mid + 1, r);
  Up(rt, l, r);
}
void upt(int rt, int l, int r, int L, int R, int val) {
  if (R < L) return;
  if (R < l || r < L) return;
  if (L <= l && r <= R) {
    T[rt].addlz(val);
    lz[rt] += val;
    return;
  }
  Down(rt, l, r);
  int mid = (l + r) / 2;
  upt((rt << 1), l, mid, L, R, val);
  upt((rt << 1 | 1), mid + 1, r, L, R, val);
  Up(rt, l, r);
}
void upt1(int rt, int l, int r, int L, int val) {
  if (l == r) {
    T1.upt(L, val);
    return;
  }
  Down(rt, l, r);
  int mid = (l + r) / 2;
  if (L <= mid)
    upt1((rt << 1), l, mid, L, val);
  else
    upt1((rt << 1 | 1), mid + 1, r, L, val);
  Up(rt, l, r);
}
void modify(int x, int sig) {
  if (pos[x].empty()) return;
  upt1(1, 1, n + 1, *pos[x].begin(), sig * pos[x].size());
  upt(1, 1, n + 1, *pos[x].begin() + 1, *pos[x].rbegin(), sig);
}
void fmain(int tid) {
  scanf("%d%d", &n, &Q);
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%d", a + i);
    pos[a[i]].insert(i);
  }
  T1.init(n);
  for (int(i) = 1; (i) <= (int)(200135 - 5); (i)++)
    if (!pos[i].empty()) {
      T1.upt(*pos[i].begin(), (int)pos[i].size());
    }
  build(1, 1, n + 1);
  for (int(i) = 1; (i) <= (int)(200135 - 5); (i)++)
    if (!pos[i].empty()) {
      upt(1, 1, n + 1, *pos[i].begin() + 1, *pos[i].rbegin(), 1);
    }
  printf("%d\n", n - T[1].sum);
  for (int(i) = 1; (i) <= (int)(Q); (i)++) {
    int j, x;
    scanf("%d%d", &j, &x);
    int val = a[j];
    a[j] = x;
    modify(val, -1);
    pos[val].erase(j);
    modify(val, 1);
    modify(x, -1);
    pos[x].insert(j);
    modify(x, 1);
    printf("%d\n", n - T[1].sum);
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}